Compactness of space of Lipschitz Continuous functionsCompactness of $(M, d)$ in terms of continuous functions $f:Mto mathbb R$Let $f$ be a continuous functions from $[0,1]$ to $mathbbR$. Then, $f$ is not necessarrily lipschitz.Compactness and Lipschitz functionsA strange criterion for compactnessCharacterizing compactness by properties of continuous functions defined on a spaceWhat is the relation between compactness , connectedness and continuous real-valued functions on $mathbbR^n$, n > 1?Does locally Lipschitz imply Lipschitz on closed balls?Are continuous mappings on a compact metric space Lipschitz?Closed & boundedness, sequentially compactness and CompletenessCompact subset of space of Continuous functions

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Compactness of space of Lipschitz Continuous functions


Compactness of $(M, d)$ in terms of continuous functions $f:Mto mathbb R$Let $f$ be a continuous functions from $[0,1]$ to $mathbbR$. Then, $f$ is not necessarrily lipschitz.Compactness and Lipschitz functionsA strange criterion for compactnessCharacterizing compactness by properties of continuous functions defined on a spaceWhat is the relation between compactness , connectedness and continuous real-valued functions on $mathbbR^n$, n > 1?Does locally Lipschitz imply Lipschitz on closed balls?Are continuous mappings on a compact metric space Lipschitz?Closed & boundedness, sequentially compactness and CompletenessCompact subset of space of Continuous functions













0












$begingroup$


Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .



What can we say about the compactness of$ (X,d).$



I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.



I can't think of anything else. Kindly help !!










share|cite|improve this question











$endgroup$











  • $begingroup$
    "A space is compact iff every continuous real function is bounded." That is not true.
    $endgroup$
    – Mars Plastic
    Mar 12 at 12:59










  • $begingroup$
    Is there a different version of result ?@MarsPlastic
    $endgroup$
    – Devendra Singh Rana
    Mar 12 at 13:04










  • $begingroup$
    A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
    $endgroup$
    – Mars Plastic
    Mar 12 at 13:06
















0












$begingroup$


Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .



What can we say about the compactness of$ (X,d).$



I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.



I can't think of anything else. Kindly help !!










share|cite|improve this question











$endgroup$











  • $begingroup$
    "A space is compact iff every continuous real function is bounded." That is not true.
    $endgroup$
    – Mars Plastic
    Mar 12 at 12:59










  • $begingroup$
    Is there a different version of result ?@MarsPlastic
    $endgroup$
    – Devendra Singh Rana
    Mar 12 at 13:04










  • $begingroup$
    A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
    $endgroup$
    – Mars Plastic
    Mar 12 at 13:06














0












0








0





$begingroup$


Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .



What can we say about the compactness of$ (X,d).$



I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.



I can't think of anything else. Kindly help !!










share|cite|improve this question











$endgroup$




Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .



What can we say about the compactness of$ (X,d).$



I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.



I can't think of anything else. Kindly help !!







general-topology functional-analysis metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 12 at 13:02







Devendra Singh Rana

















asked Mar 12 at 10:56









Devendra Singh RanaDevendra Singh Rana

7871416




7871416











  • $begingroup$
    "A space is compact iff every continuous real function is bounded." That is not true.
    $endgroup$
    – Mars Plastic
    Mar 12 at 12:59










  • $begingroup$
    Is there a different version of result ?@MarsPlastic
    $endgroup$
    – Devendra Singh Rana
    Mar 12 at 13:04










  • $begingroup$
    A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
    $endgroup$
    – Mars Plastic
    Mar 12 at 13:06

















  • $begingroup$
    "A space is compact iff every continuous real function is bounded." That is not true.
    $endgroup$
    – Mars Plastic
    Mar 12 at 12:59










  • $begingroup$
    Is there a different version of result ?@MarsPlastic
    $endgroup$
    – Devendra Singh Rana
    Mar 12 at 13:04










  • $begingroup$
    A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
    $endgroup$
    – Mars Plastic
    Mar 12 at 13:06
















$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59




$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59












$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04




$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04












$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06





$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06











1 Answer
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$begingroup$

If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.






share|cite|improve this answer









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    1 Answer
    1






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    active

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    $begingroup$

    If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.






    share|cite|improve this answer









    $endgroup$

















      4












      $begingroup$

      If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.






      share|cite|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.






        share|cite|improve this answer









        $endgroup$



        If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 12 at 11:02









        José Carlos SantosJosé Carlos Santos

        168k22132236




        168k22132236



























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